It is not a hard one, but I still learnt a good lesson on how to optimize my strategy.

My first thought was on the right track: do a O(n) scan and do swaps to make sure each swap can put one number into correct bucket. However I was trying to find the target value for the current bucket - like, when checking a[i], i was trying to look for i + 1, which is unnecessarily complex. However, since i already know value of a[i], i can put that number to its expected slot, which is much much easier.

int minimumSwaps(vector<int> arr) {
    int cnt = 0;
    for (int i = 0; i < arr.size(); i ++){
        if (arr[i] == i + 1) continue;
        swap(arr[i], arr[arr[i] - 1]);
        cnt ++;
        i --;
    }
    return cnt;
}